Rationalizing Denominator w/ a radical

Question

Reading "Art of Problem Solving : Vol. 1". Stuck on Excercise 1-6 :

$$ {\sqrt2\over \sqrt6-2} $$

I know, we must rationalize, multiplying by

$$ { \sqrt6+2\over \sqrt6+2} $$

However, what would the final product be? Thanks!

Answer 1

specifically would be:

$$\frac{\sqrt{2}}{\sqrt{6} - 2} \cdot 1 = \frac{\sqrt{2}}{\sqrt{6} - 2} \cdot \frac{\sqrt{6} + 2}{\sqrt{6} + 2} = \frac{\sqrt{2}\sqrt{6} + 2\sqrt{2}}{6 + 2\sqrt{6} - 2\sqrt{6} - 4}$$

in these last term remember that $(a + b)\cdot(a - b) = a² - ab + ab - b²$ for this reason in the denominator $({\sqrt{6} - 2}) \cdot ({\sqrt{6} + 2}) = \sqrt{6}² - 2\sqrt{6} + 2\sqrt{6} - 4 = 6 - 4 = 2$

So,

$$\frac{\sqrt{2}\sqrt{6} + 2\sqrt{2}}{6 + 2\sqrt{6} - 2\sqrt{6} - 4} = \frac{\sqrt{2\cdot6} + 2\sqrt{2}}{2} = \frac{\sqrt{12}}{2} + \frac{2\sqrt{2}}{2} = \frac{\sqrt{4\cdot3}}{2} + \sqrt{2} = \frac{\sqrt{4}\sqrt{3}}{2} + \sqrt{2} = \sqrt{3} + \sqrt{2}$$

Answer 2

$$\frac{\sqrt2}{\sqrt6-2}\cdot\frac{\sqrt6+2}{\sqrt6+2}=\frac{2\sqrt3+2\sqrt2}{2}=\sqrt3+\sqrt2$$